The velocity of a particle moving along the $x$ -axis is $v(t)=t^2+t$. At $t=1$, its position is $1$. What is the position of the particle, $s(t)$, at any time $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $s(t)=\dfrac{1}{3}t^3+\dfrac{1}{2}t^2+\dfrac{1}{6}$ (Choice B) B $s(t)=\dfrac{1}{3}t^3+\dfrac{1}{2}t^2+\dfrac{5}{6}$ (Choice C) C $s(t)=t^3+t^2-1$ (Choice D) D $s(t)=\dfrac{t^3}{3}+1$
Solution: We know that $s(t)= \int v(t) \,dt$. In this case, $s(t)= \int t^2+t \,dt$ Let's find the indefinite integral: $\begin{aligned} \int t^2+t \,dt&=\dfrac{1}{3}t^3+\dfrac{1}{2}t^2+C\\ \end{aligned}$ We know that $s(1)=1$. Let's use this information to solve for $C$. $\begin{aligned}s(t)&=\dfrac{1}{3}t^3+\dfrac{1}{2}t^2+C\\ \\ s(1)&=\dfrac{1}{3}(1)^3+\dfrac{1}{2}(1)^2+C\\ \\\\ 1&=\dfrac{1}{3}+\dfrac{1}{2}+C\\ \\ \dfrac{1}{6}&=C \end{aligned}$ The position of the particle $s(t)$ at any time $t$ is $s(t)=\dfrac{1}{3}t^3+\dfrac{1}{2}t^2+\dfrac{1}{6}$.